Labs and stuff!

We do a ton of labs in this class, which is awesome and good and delightful, but I'm too lazy to make individual posts for each.  And apparently too lazy to take pictures of the labs.  Which is something that I constantly regret...


The Unknown Element Lab


In this lab we were given a chart with moles of each beaker on it.  The beakers had A-J labels on them.  We weighed the beakers one by one to determine the amount of grams, taking into account of course the weight of the beakers, and divided that amount by the number of moles corresponding to the letter on the beaker.  It was really fun because it was easy.

The Burnt Popcorn Lab

Coat the bottom of a beaker with enough vegetable oil to just barely cover the surface.
Make a cover of aluminum for the beaker and then weigh it. (190.5g)
Add a few kernals, just one layer of kernals on the oil, and weigh it again. (207.65g)
Now situate that baby over a bunsen burner and let. It. Burn.  We did.  Pretty sure some of the popcorn caught fire too but that's how we roll.  Once all the popcorn is popped weigh it again. (205g)
Now, take the initial weight of the kernals by subtracting 190.5 from 207.65.  This gives you 17.15g.  Then, subtract 190.5 from 205 to get 14.5g, this is the weight of the popped kernal.
Divide 14.4 by 17.15 and you get the percent of the weight of the kernal without the water, 84%.  So, the rest of that percentage must be water.  Kernals of popcorn are about 16% water.  Apparently popcorn pops because of the water.  When the water heats up it bursts the kernal open and voila!
We weren't allowed to eat the popcorn we made, not because it was burned but because we used a beaker that was once a bacterial pond water growth lab.  But that didn't bother us, it's a lot easier to cook unburnt popcorn with the ancient microwave in the inner hex.

Hydrates and Predictions

You should see my notes for these labs. They're hilarious!  If I didn't know my head I'd have no idea what these stupid scribbles are.
Anyway!
The point of this lab was to figure out what percent of CuSO4x5H2O is water.
We made our predictions beforehand.
Using 5.25 g of the compound we guessed that approx. 1.5 g of it was water.
We heated the Copper Sulfate in a test tube over a bunsen burner until it ceased all blueness.  When that cooled we measured the copper sulfate again and found that we now had 3.45.  So, 1.8g of the copper sulfate stuff was water, that's about 35%.  We were pretty close but that was pure luck.  Compounds like CuSO4x5H2O can be calculated before hand to decide what percent would be water by adding the moles of all the elements together and tabling that with the grams of the compound we'd use.  It saves a lot of time if you're not as lucky as we are.

Baking Soda Lab


The goal of this lab was to get the baking soda to dissolve completely in the vinegar so that there was no longer a reaction.  Usually this takes a lot of guess work on the measurement of baking soda to use so that it's no longer reactive.  My group took the measurements of our supplies and calculated the perfect amount of baking soda to use before we even started so that we didn't have to spend time doing any guess work.  But the cat got my data for this lab so I don't have the exact numbers.  Not that they were important unless you need the perfect ratio of vinegar to baking soda for something practical.
All we did was take the amount of vinegar and cross multiply with an x amount of baking soda.  It was pretty basic algebra, a part of algebra that hasn't been replaced by my obviously practical need to be a calculus whiz.  The x amount of baking soda was the amount we needed to achieve that 1:1 ratio.  Worked like a dream.

The Mr. Ludwig's Retirement Fund Lab

Or the silver copper replacement lab.
This lab took a couple of days to complete.
We took 30cm of copper (3.49g) and coiled the end loosely so it would fit inside a tiny test tube.  We added distilled water and 1g of silver nitrate to the beaker then dropped in the copper and sealed it.  The copper dulled instantly and shaking the tube lightly made the copper turn dark as the silver took hold.  Then we just left it alone for a day.  Or over the weekend?  I can't remember, I was bed ridden for a few days.
Apparently my partner had to spray the silver off of the copper with distilled water and into a funnel made of filter paper.  That sat for a day as well to let the water filter through completely.  After this we weighed the silver and copper again.  The .35g of silver went to Mr. Ludwig's Retirement Fund, he'll be set.  And the 3.193g of copper went in the trash or something.  All we needed were the numbers.
Our stoichometrically predicted number of how much silver would be produced from this lab was .65g.  This was quite wrong obviously so perhaps we messed up the math a bit.  I'd rather blame the method of silver extraction though.  When shaking the silver from the filter paper the paper was quite gray so obviously that didn't come away clean and this is most likely why our numbers didn't match up.  I like that answer a lot better than wrong math.

So that's it for all the labs!  Sorry there aren't more pictures.  I'll remember to take lots of pictures for all the labs next quarter.

Moles of Information!

See Mr. Ludwig?  I pay attention to your lecture slides!

The beginning of this quarter had a lot to do with moles.  Moles are the system used for measuring elements.  Each and every element on the periodic table has a different weight (in grams) to equal one mole.  One mole is 6.02x10^23, 6 billion trillion, Amadeo Avogadro's number.  That's how many molecules it takes an element to equal a mole.  In any other form but molecules that number is so large it would swamp us.  But when dealing with elements, it's perfect for making any measurement or equation we would need to solve a problems like how much of an element there is in a mixture and such.
But how can we tell how much of an element we need in order to have 1 mole of that element?  It's simple, the number is on the periodic table with the atom!  It's the decimal number by the abbreviation.  For example, the grams of Iron it takes to have 1 mole of iron is 55.845g.  For oxygen it's 15.999g.  Easy, and readily available through the work of other people! Yay!!

Mole Equations

Now we know what moles are and how to figure out how many g/mole are in a single element.  But what about compound elements?
Let's take a look at H2O.
That's 2 hydrogens for every oxygen.  So, to get the g/mole of H2O we simply add all of the elements together.
2 moles of hydrogen x 1.0079 + 1 mole of oxygen x 15.999 = 18.015 g/mole of H2O.
Alternately, you can find out how many moles are in, let's say, 262g of H2O.  Just take the H2O and calculate it to g/mole.  Then, we'll divide the 262 grams of H2O we have by that base number of 18.015.  Doing so gives us about 14.543 moles in 262 grams of H2O.  Easy!
Let's start with a number of moles now.  How  many grams of H2O are in 23 moles?  Again, you would add all the elements in H2O together to get a base number to multiply by.  Then we take that number and multiply it by the moles.
23 x 18.015 = 414.385 grams in 23 moles of H2O
I use H2O as my example because we used this compound a lot in our equations and calculations in class.

Believe it or not, this gets a little more complicated.


Not sure why I'd ever need this but through moles we can also find out how many atoms are in grams of an element.
Let's do this with 35.4g of Cu.
First, we've got to get out of the g part of the equation.
So we'll set it up in a way that we can do that.
35.4g Cu | 1 mole Cu | 6.02 x 10^23 atoms Cu
----------------------------------------------
               | 63.4g Cu   |        1 mole Cu
In English we're multiplying 35.4 by 1/63.4 by 6.02 x 10^23/1 to get 3.4 x 10^23 atoms Cu.  This way, we cancel out the unites of moles and the units of grams to end up with just the units of atoms.  And I use Mr. Ludwigs example for this equation because I always, without fail, mess up powers of 10 calculations, somehow.  I could probably do this exact same equation perfectly with x and pi instead of actual numbers but the basic algebra and math has been trained out of me by upper level mathematics.  But that's high school for you.  Pass until you fail.

There are three formulas for finding the atoms in a compound or in grams.  Someone kill me if I ever have to care about how many atoms are in 23g of vanadium.  Or any other element for that matter.  (Haha matter.)
The Empirical Formula calculates the smallest whole number ratio of atoms.
Ionic formulas are Empirical.
The Molecular Formula gives the number of each kind of atom in one molecule of a compound.
I think I did some of these in my notebook.  I'm looking at a bunch of scribbles now that look like they could be this.  Wanna see my test?  Well you can't because my cat ripped it to shreds.  With absolutely no encouragement from me.
I'm a reader, dudes, not a counter!!!